# A question related to the method of images and uniqueness theorems

1. Sep 27, 2015

### davidbenari

My question is best illustrated by an example from a Griffiths book on E&M:

"A point charge q is situated a distance $a$ from the center of a grounded conducting sphere of radius R ($a>R$). Find the potential outside the sphere... With the addition of a second charge you can simulate any $V_0$ on the sphere now... Solve this problem now with a neutral conducting sphere at potential $V_0$."

Now, this problem is pretty well-known and it can be shown that adding a charge $q'=-\frac{R}{a}q$ a distance $b=\frac{R^2}{a}$ collinear with the origin of the sphere and the original charge $q$, solves the boundary problem for a grounded sphere. If I want to create a new boundary voltage $V_0$ it is simple to recognize that I have to put a second image charge $q''$ on the origin such that $V_0=\frac{kq''}{R}$.

But I'm confused as to why for a neutral conducting sphere my two image charges have to satisfy $q''+q'=0$. Uniqueness theorems talk about boundary conditions, and in a way I only have to satisfy a specified voltage $V_0$ on the boundaries, why do I also have to satisfy the conditions of charge inside this sphere? What part of the uniqueness theorem say so?

I know the uniqueness theorems talk about specifying voltage or net charge on a boundary. But when solving a PDE giving both net charge and voltage on a conducting surface usually overdetermines the system, and talking about net charge gives you information about a surface integral which I think is pretty useless information due to the complexity of such a condition.

For anyone not acquainted with this problem, I think they state it clearly on this website:http://ocw.nctu.edu.tw/upload/classbfs120903470749199.pdf

Last edited by a moderator: Sep 27, 2015
2. Sep 28, 2015

### lautaaf

The problem we are looking to solve it is: $$\Delta \Phi = - \frac {\rho}{\epsilon_0} \qquad \qquad \text{ in } \Omega_{out}=\{\vec x \in \mathbb R^{3} : \|\vec x \| > R\} \quad (1) \\ \Phi \big|_{r=R} =0 \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \: \: \quad \quad(2) \\ \lim_{r\rightarrow +\infty} \Phi = 0 \qquad \qquad \quad\qquad\qquad \qquad \qquad \qquad \qquad \quad (3)$$.

The method of the images takes advantage of the following: You can "do" whatever you like inside the $\Omega_{in}$ region because it's not part of your problem. In particular, you can simulate the boundary condition placing a image charge inside $\Omega_{in}$. That's gonna help you find a potencial $\Phi$ (defined on $\Omega_{out}$) that satisfies $(1),(2)\text { & }(3)$.
The condition $q' + q \frac{R}{a}=0$ (check the pdf) is the relation $q$ and $q'$ must obey in order to have $V=0$ on the boundary. This is not an additional condition, it's a consequence of asking $V=0$ on the boundary.
The case of a sphere at potential $V_0 \neq 0$ is similar: since $(1)$ is a linear eq, you can use the superposition principle, and consider the problem as a sum of two problems: a charge situated a distance $a$ from the center of a grounded sphere and a isolated sphere at potential $V_0$

3. Sep 29, 2015

### ehild

In case of a grounded sphere, a nearby point charge q repel or attract charges from the ground, and the sphere becomes charged. That charge distribution can be replaced by the image charge.
If the sphere is not grounded, and was neutral initially, it must stay neutral when the external charge is placed nearby. There is charge redistribution on the sphere, but the the net charge stays zero. Introducing the image charge brings a new charge into the set-up which has to be balanced. That is an opposite point charge at the centre of the sphere.

4. Sep 29, 2015

### davidbenari

Yes but when the sphere was neutral initially I also have to take this into account when placing image charges and my question was why? Uniqueness theorems don't say anything about this.

How did you know it had to be balanced? Uniqueness theorems don't speak about this. What I had in mind as a justification was Gauss's law. The surface integral of the original problem should be zero, and the image problem should also be zero if it is to match the real problem... but I'm not so sure this is correct.

5. Sep 29, 2015

### ehild

Just because of Gauss' Law. If the sphere is neutral, and it is not grounded, so no charge can leave it, no charge can go on it, the flux through a Gaussian surface enclosing the sphere but not the charge near it, should be zero, as the enclosed charge is zero.
If you replace the sphere with the image charge, the flux through the same Gaussian surface as before will indicate enclosed charge, which is not there in reality. So you need an other charge to make the sphere neutral.